Equivariant singular cohomology - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:06:10Z http://mathoverflow.net/feeds/question/9581 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9581/equivariant-singular-cohomology Equivariant singular cohomology Kevin Lin 2009-12-23T02:30:32Z 2011-02-25T04:35:52Z <p>One can define the $G$-equivariant cohomology of a space $X$ as being the ordinary singular cohomology of $X \times_G EG$ --- I think this is due to Borel? (See e.g. section 2 of <a href="http://arxiv.org/abs/0709.3615" rel="nofollow">these notes</a>)</p> <p>Alternatively if $X$ is a manifold, we also have $G$-equivariant de Rham cohomology, defined in terms of $G$-equivariant differential forms --- I think this is due to Cartan? (See e.g. section 3 of loc. cit.)</p> <p>I suspect this is extremely standard or obvious, but if it is, I don't know where it's written down: Is it possible to define equivariant cohomology of a topological space in terms of some notion of "equivariant singular cochains", that is, <em>without</em> using the Borel construction?</p> http://mathoverflow.net/questions/9581/equivariant-singular-cohomology/9583#9583 Answer by Kevin McGerty for Equivariant singular cohomology Kevin McGerty 2009-12-23T03:02:16Z 2009-12-23T03:02:16Z <p>Here's an answer which I learned from Goresky-Kottwitz-MacPherson's paper on equivariant cohomology and Koszul duality: they use some notion of geometric chain which is probably something like subanalytic chains, but anyway, the idea is as follows.</p> <p>Suppose $G$ is a compact Lie group of dimension d. An abstract equivariant $k$-chain $c$ is a $(k+d)$-dimensional chain in some $\mathbb R^n$ (or perhaps it's better to say $\mathbb R^\infty$) equipped with a free action of $G$. Then if $X$ is a $G$-space, an equivariant chain in $X$ is a $K$-equivariant map from an abstract chain to $X$. You can obviously form a chain complex out of these things, and the result gives you the equivariant cohomology of $X$ (in the Borel construction say).</p> http://mathoverflow.net/questions/9581/equivariant-singular-cohomology/9588#9588 Answer by Ben Webster for Equivariant singular cohomology Ben Webster 2009-12-23T03:57:23Z 2011-02-24T22:08:29Z <p>You can think of it as the cohomology of the simplicial manifold $X\leftleftarrows X\times G \cdots$ where the $n$-simplices are $X\times G^n$ and the face maps either act on $X$ or multiply two consecutive entries. </p> <p>Of course, some people will tell you that that is really the same as the Borel construction, but if you're willing to interpret things that liberally, you'll never get away from the Borel construction.</p> http://mathoverflow.net/questions/9581/equivariant-singular-cohomology/56573#56573 Answer by Chris Gerig for Equivariant singular cohomology Chris Gerig 2011-02-24T22:44:57Z 2011-02-24T22:44:57Z <p>This isn't singular, but in the same spirit as what you want (i.e. not the Borel construction):</p> <p>Take the cellular chain complex $C(X)$ of a $G$-complex $X$. You can define cohomology groups with coefficients in a chain complex, and $H^*(G,C(X))$ is defined to be the equivariant cohomology of $X$ (for $X$ finite-dimensional and $G$ finite).</p> <p>This is explained in Ken Brown's $\textit{Cohomology of Groups}$ (chapter VII).</p> http://mathoverflow.net/questions/9581/equivariant-singular-cohomology/56595#56595 Answer by Peter May for Equivariant singular cohomology Peter May 2011-02-25T04:35:52Z 2011-02-25T04:35:52Z <p>In 1965 or so, Glen Bredon defined ordinary equivariant cohomology, ordinary meaning that it satisfies the dimension axiom: For each coefficient system $M$ (contravariant functor from the orbit category of G to the category of Abelian groups), there is a unique cohomology theory $H^*_G(-;M)$ such that, when restricted to the orbit category, it spits out the functor $M$. Just as in the nonequivariant world, it can be defined using either singular or cellular cochains, the latter defined using $G$-CW complexes. This works as stated for any topological group $G$.</p> <p>For an abelian group $A$, Borel cohomology with coefficients in $A$, $H^*(EG\times_G X;A)$ is the extremely special case in which one takes $M$ to be the constant coefficient system $\underline{A}$ at the group $A$ and replaces $X$ by $EG\times X$. That is,</p> <p>$$H^*_G(EG\times X; \underline{A}) = H^*(EG\times_G X;A)$$ </p>